3.14 \(\int \frac{(e x)^m (A+B x^2) (c+d x^2)^2}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=292 \[ -\frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (a d (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))-b c (a B (m+1)+A b (3-m)) (a d (m+1)+b (c-c m)))}{8 a^3 b^3 e (m+1)}+\frac{(e x)^{m+1} (b c-a d) \left (c (a B (m+1)+A b (3-m))-d x^2 (A b (m+1)-a B (m+5))\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac{d (e x)^{m+1} (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))}{8 a^2 b^3 e (m+1)}+\frac{\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \]

[Out]

(d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(1 + m) - a*B*(5 + m))*(e*x)^(1 + m))/(8*a^2*b^3*e*(1 + m)) + ((A*b - a*B)
*(e*x)^(1 + m)*(c + d*x^2)^2)/(4*a*b*e*(a + b*x^2)^2) + ((b*c - a*d)*(e*x)^(1 + m)*(c*(A*b*(3 - m) + a*B*(1 +
m)) - d*(A*b*(1 + m) - a*B*(5 + m))*x^2))/(8*a^2*b^2*e*(a + b*x^2)) - ((a*d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(
1 + m) - a*B*(5 + m)) - b*c*(A*b*(3 - m) + a*B*(1 + m))*(a*d*(1 + m) + b*(c - c*m)))*(e*x)^(1 + m)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*b^3*e*(1 + m))

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Rubi [A]  time = 0.408485, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {577, 459, 364} \[ -\frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (a d (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))-b c (a B (m+1)+A b (3-m)) (a d (m+1)+b (c-c m)))}{8 a^3 b^3 e (m+1)}+\frac{(e x)^{m+1} (b c-a d) \left (c (a B (m+1)+A b (3-m))-d x^2 (A b (m+1)-a B (m+5))\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac{d (e x)^{m+1} (A b (m+1)-a B (m+5)) (b c (m+1)-a d (m+3))}{8 a^2 b^3 e (m+1)}+\frac{\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(A + B*x^2)*(c + d*x^2)^2)/(a + b*x^2)^3,x]

[Out]

(d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(1 + m) - a*B*(5 + m))*(e*x)^(1 + m))/(8*a^2*b^3*e*(1 + m)) + ((A*b - a*B)
*(e*x)^(1 + m)*(c + d*x^2)^2)/(4*a*b*e*(a + b*x^2)^2) + ((b*c - a*d)*(e*x)^(1 + m)*(c*(A*b*(3 - m) + a*B*(1 +
m)) - d*(A*b*(1 + m) - a*B*(5 + m))*x^2))/(8*a^2*b^2*e*(a + b*x^2)) - ((a*d*(b*c*(1 + m) - a*d*(3 + m))*(A*b*(
1 + m) - a*B*(5 + m)) - b*c*(A*b*(3 - m) + a*B*(1 + m))*(a*d*(1 + m) + b*(c - c*m)))*(e*x)^(1 + m)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*b^3*e*(1 + m))

Rule 577

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*b*g*n*(p + 1)), x] + Dist[
1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(m
+ 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] &&
IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^3} \, dx &=\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}-\frac{\int \frac{(e x)^m \left (c+d x^2\right ) \left (-c (A b (3-m)+a B (1+m))+d (A b (1+m)-a B (5+m)) x^2\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac{(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac{\int \frac{(e x)^m \left (c (A b (3-m)+a B (1+m)) (b c (1-m)+a d (1+m))+d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) x^2\right )}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac{d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac{(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac{\left (\frac{a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))}{b}-c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))\right ) \int \frac{(e x)^m}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac{d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac{(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac{(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac{\left (\frac{a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))}{b}-c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))\right ) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{8 a^3 b^2 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.170024, size = 165, normalized size = 0.57 \[ \frac{x (e x)^m \left (\frac{(b c-a d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (-3 a B d+2 A b d+b B c)}{a^2}+\frac{(A b-a B) (b c-a d)^2 \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^3}+\frac{d \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (-3 a B d+A b d+2 b B c)}{a}+B d^2\right )}{b^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(A + B*x^2)*(c + d*x^2)^2)/(a + b*x^2)^3,x]

[Out]

(x*(e*x)^m*(B*d^2 + (d*(2*b*B*c + A*b*d - 3*a*B*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/a
+ ((b*c - a*d)*(b*B*c + 2*A*b*d - 3*a*B*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/a^2 + ((
A*b - a*B)*(b*c - a*d)^2*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/a^3))/(b^3*(1 + m))

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( B{x}^{2}+A \right ) \left ( d{x}^{2}+c \right ) ^{2}}{ \left ( b{x}^{2}+a \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x)

[Out]

int((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (d x^{2} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(d*x^2 + c)^2*(e*x)^m/(b*x^2 + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B d^{2} x^{6} +{\left (2 \, B c d + A d^{2}\right )} x^{4} + A c^{2} +{\left (B c^{2} + 2 \, A c d\right )} x^{2}\right )} \left (e x\right )^{m}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

integral((B*d^2*x^6 + (2*B*c*d + A*d^2)*x^4 + A*c^2 + (B*c^2 + 2*A*c*d)*x^2)*(e*x)^m/(b^3*x^6 + 3*a*b^2*x^4 +
3*a^2*b*x^2 + a^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x**2+A)*(d*x**2+c)**2/(b*x**2+a)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (d x^{2} + c\right )}^{2} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)*(d*x^2+c)^2/(b*x^2+a)^3,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(d*x^2 + c)^2*(e*x)^m/(b*x^2 + a)^3, x)